Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, i, j, k, l, coun = 0, coun1 = 0;
  cin >> n >> m;
  int a[n][m], b[n], c[m], d[n], f[m], e[n][m];
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
      e[i][j] = a[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    int ma = 1000;
    for (j = 0; j < m; j++) {
      ma = min(a[i][j], ma);
    }
    b[i] = ma;
    coun += ma;
    for (j = 0; j < m; j++) {
      a[i][j] -= ma;
    }
  }
  for (j = 0; j < m; j++) {
    int ma = 1000;
    for (i = 0; i < n; i++) {
      ma = min(ma, a[i][j]);
    }
    c[j] = ma;
    coun += ma;
    for (i = 0; i < n; i++) {
      a[i][j] -= ma;
    }
  }
  int flag = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] > 0) {
        flag = 1;
        break;
      }
    }
  }
  if (flag) {
    cout << -1 << endl;
  } else {
    for (j = 0; j < m; j++) {
      int ma = 1000;
      for (i = 0; i < n; i++) {
        ma = min(ma, e[i][j]);
      }
      f[j] = ma;
      coun1 += ma;
      for (i = 0; i < n; i++) {
        e[i][j] -= ma;
      }
    }
    for (i = 0; i < n; i++) {
      int ma = 1000;
      for (j = 0; j < m; j++) {
        ma = min(e[i][j], ma);
      }
      d[i] = ma;
      coun1 += ma;
      for (j = 0; j < m; j++) {
        e[i][j] -= ma;
      }
    }
    k = min(coun, coun1);
    cout << k << endl;
    if (k == coun) {
      for (i = 0; i < n; i++) {
        if (b[i] > 0) {
          for (j = 0; j < b[i]; j++) {
            cout << "row " << i + 1 << endl;
          }
        }
      }
      for (j = 0; j < m; j++) {
        if (c[j] > 0) {
          for (i = 0; i < c[j]; i++) {
            cout << "col " << j + 1 << endl;
          }
        }
      }
    } else {
      for (i = 0; i < n; i++) {
        if (d[i] > 0) {
          for (j = 0; j < d[i]; j++) {
            cout << "row " << i + 1 << endl;
          }
        }
      }
      for (j = 0; j < m; j++) {
        if (f[j] > 0) {
          for (i = 0; i < f[j]; i++) {
            cout << "col " << j + 1 << endl;
          }
        }
      }
    }
  }
}